## Calculus (3rd Edition)

$f(x)=\frac{\cos x^2}{x^2-1}$ is continuous on $R- \{-1,1\}$.
Since $x^2-1=0$, then $x=\pm 1$. Then, $f(x)=\frac{\cos x^2}{x^2-1}$ is defined for all $x\in R\backslash \{-1,1\}$. Now, since $x^2-1$ and $\cos x^2$ are continuous, then by using the continuity law $f(x)=\frac{\cos x^2}{x^2-1}$ is continuous on $R- \{-1,1\}$.